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TheoremStatement: AI-adaptedProof: AI-generatedPipeline-generated
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  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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The computational basis is orthonormal

Statement

For every d≥1, the ordered computational basis (e0,…,ed−1) of the register Qd from Finite registers and their ordered computational bases is an orthonormal basis for its declared inner product. Separately, the one-dimensional zero-qubit register has the basis {1} by definition.

Facts & Assumptions

Given: A positive integer d, the coordinate space Cd, and its standard pairing.

[F1]

The standard pairing on Cd is ⟨z,w⟩root=∑r<dzrwr‾ and is linear in the first argument (The standard formulas ⟨x,y⟩=∑k<nxkyk on Rn and ∑k<nxkyk‾ on Cn are inner products).

[F3]

The register Qd uses this ordered standard basis and this pairing. Separately, the zero-qubit register is the empty tensor product C with basis {1}, distinct from C0 (Finite registers and their ordered computational bases).

Proof

1.1givenF1F3algebra

For j,k<d, the coordinate formula in [F1] gives ⟨ej,ek⟩root=∑r<dδrjδrk, which equals 1 when j=k and 0 when j≠k. Thus the listed vectors are orthonormal.

2.1step 1.1F2F3

By [F2] the same vectors form a basis, so they are an orthonormal basis of Qd.

3.1F3∎

By [F3], the zero-qubit register is the empty tensor product C with basis {1}, distinct from the zero-dimensional coordinate space C0.

Depends on

Used by

…and 70 more results.

Cited to discharge well-definedness by Finite registers and their ordered computational bases.

Dependency tree · two levels

29 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources